Julian Zbigniew Kuryllowicz-Kazmierczak has authored 3 sequences.
A370399
Triangle read by rows: T(n, k) is the denominator of the probability of winning a certain game while playing optimally.
Original entry on oeis.org
1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 18, 6, 1, 1, 15, 60, 15, 10, 1, 1, 15, 2, 100, 25, 20, 1, 1, 15, 210, 75, 1050, 105, 35, 1, 1, 15, 5880, 5880, 73500, 29400, 588, 70, 1, 1, 135, 30240, 35280, 529200, 1852200, 15435, 588, 126, 1, 1, 135, 340200, 453600, 7938000, 466754400, 3111696, 1481760, 19845, 252, 1
Offset: 1
The values of Pr(n,k) begin as follows:
.
n\k| 0 1 2 3 4 5 6 7
---+---------------------------------------------------------
1 | 0/1 1/1
2 | 0/1 1/2 1/1
3 | 0/1 1/3 2/3 1/1
4 | 0/1 1/3 11/18 5/6 1/1
5 | 0/1 4/15 31/60 11/15 9/10 1/1
6 | 0/1 4/15 1/2 69/100 21/25 19/20 1/1
7 | 0/1 4/15 101/210 49/75 829/1050 94/105 34/35 1/1
...
We can calculate Pr(4,2) using the table below, given the values of Pr(n,k) for n=3 and for n=2. The leftmost column lists each of the six possible results of randomizing the n=4 tokens during the first move; in each randomized sequence, the red and blue tokens are represented by "r" and "b", respectively.
.
randomized probability result if result if
sequence of last 1 token last 2 tokens
of tokens occurrence is removed are removed
========== =========== ============== =============
rrbb 1/6 Pr(3,1) = 1/3 Pr(2,0) = 0/1
rbrb 1/6 Pr(3,1) = 1/3 Pr(2,1) = 1/2
brrb 1/6 Pr(3,1) = 1/3 Pr(2,1) = 1/2
rbbr 1/6 Pr(3,2) = 2/3 Pr(2,1) = 1/2
brbr 1/6 Pr(3,2) = 2/3 Pr(2,1) = 1/2
bbrr 1/6 Pr(3,2) = 2/3 Pr(2,2) = 1/1
.
For example, when we get rbrb it's better to remove the last two tokens (one r and one b) instead of removing only the last token (b). So the probability of winning M(4,2) is
Pr(4,2) = (1/6)(1/3) + (1/6)(1/2) + (1/6)(1/2) + (1/6)(2/3) + (1/6)(2/3) + (1/6)(1/1) = 11/18.
Of course Pr(n,k) >= k/n, because k/n could be achieved by removing 1 token on each move.
A370398
Triangle read by rows: T(n,k) is the numerator of the probability of winning a 1-player game M(n,k) as defined below while playing optimally.
Original entry on oeis.org
0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 11, 5, 1, 0, 4, 31, 11, 9, 1, 0, 4, 1, 69, 21, 19, 1, 0, 4, 101, 49, 829, 94, 34, 1, 0, 4, 2783, 3733, 56069, 25367, 551, 69, 1, 0, 32, 13439, 21517, 389573, 1543163, 14011, 565, 125, 1, 0, 32, 149621, 271643, 5709959, 379562191, 2757715, 1392901, 19388, 251, 1
Offset: 1
The values of Pr(n,k) begin as follows:
.
n\k| 0 1 2 3 4 5 6 7
---+---------------------------------------------------------
1 | 0/1 1/1
2 | 0/1 1/2 1/1
3 | 0/1 1/3 2/3 1/1
4 | 0/1 1/3 11/18 5/6 1/1
5 | 0/1 4/15 31/60 11/15 9/10 1/1
6 | 0/1 4/15 1/2 69/100 21/25 19/20 1/1
7 | 0/1 4/15 101/210 49/75 829/1050 94/105 34/35 1/1
...
We can calculate Pr(4,2) given the values of Pr(n,k) for n=3 and n=2 as seen in the table below. The leftmost column lists each of the six possible outcomes (i.e., C(4,2) = 6 combinations, all equally likely) of randomizing the n=4 tokens during the first move; in each randomized sequence (i.e., combination), the red and blue tokens are represented by "r" and "b", respectively. Removing the last j = 1 or 2 tokens will leave n' = n - j remaining tokens of which k' = k - f(c,j) are blue. For each randomized sequence, an asterisk marks the probability of winning using the optimal choice of the number j of tokens to remove.
.
randomized if remove if remove probability
sequence last j=1 token last j=2 tokens of win
of --------------- --------------- given optimal
tokens n' k' Pr(n',k') n' k' Pr(n',k') choice
========== == == ========= == == ========= =============
rrbb 3 1 1/3 * 2 0 0/1 1/3
rbrb 3 1 1/3 2 1 1/2 * 1/2
brrb 3 1 1/3 2 1 1/2 * 1/2
rbbr 3 2 2/3 * 2 1 1/2 2/3
brbr 3 2 2/3 * 2 1 1/2 2/3
bbrr 3 2 2/3 2 2 1/1 * 1/1
.
For example, when we get rbrb it's better to remove the last two tokens (one r and one b) instead of removing only the last token (b).
The probability of winning M(4,2) is the average of the probabilities of winning for each randomized sequence, i.e.,
Pr(4,2) = (1/3 + 1/2 + 1/2 + 2/3 + 2/3 + 1/1)/6 = 11/18.
A363245
Lexicographically first sequence of positive integers such that all terms are pairwise coprime and no subset sum is a power of 2.
Original entry on oeis.org
3, 7, 10, 11, 17, 31, 41, 71, 169, 199, 263, 337, 367, 1553, 2129, 2287, 2297, 4351, 10433, 16391, 16433, 34829, 65543, 69557, 165887, 262151, 358481, 817153, 952319, 1048583, 3704737, 3932167, 4518071, 12582919, 17305417, 17367019, 50069497, 50593799, 87228517
Offset: 1
-
a = {3}; k = 2; Monitor[Do[While[Or[! Apply[CoprimeQ, Join[a, {k}]], AnyTrue[Map[Log2 @* Total@ Append[#, k] &, Subsets[a]], IntegerQ]], k++]; AppendTo[a, k]; k++, {i, 16}], {i, k}]; a (* Michael De Vlieger, Jun 14 2023 *)
-
from math import gcd
from itertools import count, islice
def agen(): # generator of terms
a, ss, pows2, m = [], set(), {1, 2}, 2
for k in count(1):
if k in pows2: continue
elif k > m: m <<= 1; pows2.add(m)
if any(p2-k in ss for p2 in pows2): continue
if any(gcd(ai, k) != 1 for ai in a): continue
a.append(k); yield k
ss |= {k} | {k+si for si in ss if k+si not in ss}
while m < max(ss): m <<= 1; pows2.add(m)
print(list(islice(agen(), 30))) # Michael S. Branicky, Jun 09 2023
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